A ball is projected from at an angle `45^(@)` with horizontal from distance `d_(1)` from the foot of a pole and just after touching the top of pole it the falls on ground at distance `d_(2)` from pole on other side, the height of pole is

Options

A. 2`sqrt(d_(1))d_(2)`
B. `(d_(1) + d_(2))/(4)`
C. `(2 d_(1) d_(2))/(d_(1) + d_(2))`
D. `(d_(1) d_(2))/(d_(1) + d_(2))`

Answer

To solve the problem of finding the height of the pole when a ball is projected at an angle of \(45^\circ\) and travels distances \(d_1\) and \(d_2\) before and after touching the top of the pole, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Projectile Motion**:
The ball is projected at an angle of \(45^\circ\) with the horizontal. The horizontal and vertical components of the initial velocity \(u\) can be expressed as:
\[ u_x = u \cos(45^\circ) = \frac{u}{\sqrt{2}}, \quad u_y = u \sin(45^\circ) = \frac{u}{\sqrt{2}} \]

2. **Horizontal Motion**:
The horizontal distance covered when the ball reaches the height of the pole is \(d_1\). The time \(t_1\) taken to reach this point can be expressed as:
\[ d_1 = u_x t_1 \Rightarrow t_1 = \frac{d_1 \sqrt{2}}{u} \]

3. **Vertical Motion at the Height of the Pole**:
At the height of the pole \(h\), the vertical position can be described by the equation:
\[ h = u_y t_1 - \frac{1}{2} g t_1^2 \]
Substituting for \(u_y\) and \(t_1\):
\[ h = \frac{u}{\sqrt{2}} \cdot \frac{d_1 \sqrt{2}}{u} - \frac{1}{2} g \left(\frac{d_1 \sqrt{2}}{u}\right)^2 \]
Simplifying this gives:
\[ h = d_1 - \frac{g d_1^2}{2u^2} \]

4. **Horizontal Motion After the Pole**:
After touching the top of the pole, the ball travels a distance \(d_2\) before hitting the ground. The total horizontal distance from the point of projection to the point of impact is \(d_1 + d_2\). The time \(t_2\) taken to reach the ground can be expressed as:
\[ d_1 + d_2 = u_x t_2 \Rightarrow t_2 = \frac{(d_1 + d_2) \sqrt{2}}{u} \]

5. **Vertical Motion When Hitting the Ground**:
At the ground level (height = 0), we can use the vertical motion equation:
\[ 0 = u_y t_2 - \frac{1}{2} g t_2^2 \]
Substituting for \(u_y\) and \(t_2\):
\[ 0 = \frac{u}{\sqrt{2}} \cdot \frac{(d_1 + d_2) \sqrt{2}}{u} - \frac{1}{2} g \left(\frac{(d_1 + d_2) \sqrt{2}}{u}\right)^2 \]
Simplifying gives:
\[ 0 = (d_1 + d_2) - \frac{g (d_1 + d_2)^2}{2u^2} \]

6. **Solving for \(u^2\)**:
Rearranging the equation from step 5 gives:
\[ g (d_1 + d_2)^2 = 2u^2 (d_1 + d_2) \Rightarrow u^2 = \frac{g (d_1 + d_2)}{2} \]

7. **Substituting \(u^2\) back into the height equation**:
Now, substitute \(u^2\) from step 6 into the height equation from step 3:
\[ h = d_1 - \frac{g d_1^2}{2 \cdot \frac{g (d_1 + d_2)}{2}} = d_1 - \frac{d_1^2}{d_1 + d_2} \]
Simplifying this gives:
\[ h = \frac{d_1(d_1 + d_2) - d_1^2}{d_1 + d_2} = \frac{d_1 d_2}{d_1 + d_2} \]

### Final Result:
The height of the pole \(h\) is given by:
\[ h = \frac{d_1 d_2}{d_1 + d_2} \]